`MnO_(4)^(-)+NO_(2)^(-)+H^(+)rarrMn^(2+)+NO_(3)^(-)+H_(2)O`
When this equation is balanced correctly this equation is balanced with the smallest integer coefficients ?

Options

A. 1
B. 6
C. 8
D. 16

Answer

To balance the redox reaction \( \text{MnO}_4^{-} + \text{NO}_2^{-} + \text{H}^{+} \rightarrow \text{Mn}^{2+} + \text{NO}_3^{-} + \text{H}_2\text{O} \) using the half-reaction method, follow these steps:

### Step 1: Identify the oxidation and reduction half-reactions
- The manganese in \( \text{MnO}_4^{-} \) is reduced to \( \text{Mn}^{2+} \).
- The nitrogen in \( \text{NO}_2^{-} \) is oxidized to \( \text{NO}_3^{-} \).

### Step 2: Write the half-reactions
1. **Reduction half-reaction:**
\[ \text{MnO}_4^{-} \rightarrow \text{Mn}^{2+} \]
2. **Oxidation half-reaction:**
\[ \text{NO}_2^{-} \rightarrow \text{NO}_3^{-} \]

### Step 3: Balance the reduction half-reaction
- Balance Mn: already balanced.
- Balance O by adding water:
\[ \text{MnO}_4^{-} + 4 \text{H}_2\text{O} \rightarrow \text{Mn}^{2+} \]
- Balance H by adding \( \text{H}^{+} \):
\[ \text{MnO}_4^{-} + 4 \text{H}_2\text{O} \rightarrow \text{Mn}^{2+} + 8 \text{H}^{+} \]
- Balance charge by adding electrons:
\[ \text{MnO}_4^{-} + 8 \text{H}^{+} + 5 \text{e}^{-} \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

### Step 4: Balance the oxidation half-reaction
- Balance N: already balanced.
- Balance O by adding water:
\[ \text{NO}_2^{-} + \text{H}_2\text{O} \rightarrow \text{NO}_3^{-} \]
- Balance H by adding \( \text{H}^{+} \):
\[ \text{NO}_2^{-} + \text{H}_2\text{O} \rightarrow \text{NO}_3^{-} + 2 \text{H}^{+} \]
- Balance charge by adding electrons:
\[ \text{NO}_2^{-} + \text{H}_2\text{O} \rightarrow \text{NO}_3^{-} + 2 \text{H}^{+} + 2 \text{e}^{-} \]

### Step 5: Equalize the number of electrons
- Multiply the oxidation half-reaction by 5 to equalize the electrons:
\[ 5 \text{NO}_2^{-} + 5 \text{H}_2\text{O} \rightarrow 5 \text{NO}_3^{-} + 10 \text{H}^{+} + 10 \text{e}^{-} \]

### Step 6: Combine the half-reactions
- Add the balanced half-reactions:
\[ \text{MnO}_4^{-} + 8 \text{H}^{+} + 5 \text{e}^{-} + 5 \text{NO}_2^{-} + 5 \text{H}_2\text{O} \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} + 5 \text{NO}_3^{-} + 10 \text{H}^{+} + 10 \text{e}^{-} \]

### Step 7: Simplify the equation
- Cancel out the common terms:
\[ \text{MnO}_4^{-} + 5 \text{NO}_2^{-} + 6 \text{H}^{+} \rightarrow \text{Mn}^{2+} + 5 \text{NO}_3^{-} + 4 \text{H}_2\text{O} \]

### Final Balanced Equation:
\[ \text{MnO}_4^{-} + 5 \text{NO}_2^{-} + 6 \text{H}^{+} \rightarrow \text{Mn}^{2+} + 5 \text{NO}_3^{-} + 4 \text{H}_2\text{O} \]

### Coefficient for \( \text{H}^{+} \):
The coefficient for \( \text{H}^{+} \) in the balanced equation is **6**.

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